12.5 gm of fuming `H_(2)SO_(4)` (labelled as 112%) is mixed with 100 lit water. Molar concentration of `H^(+)` in resultant solution is
[Note: Assume that `H_(2)SO_(4)` dissociate completely and there is no change in volume on mixing

Options

A. `(2)/(700)`
B. `(2)/(350)`
C. `(3)/(350)`
D. `(3)/(700)`

Answer

To solve the problem step by step, we will follow the instructions provided in the video transcript while adding clarity and detail to each step.

### Step 1: Calculate the mass of `H₂SO₄` in the fuming solution
Given that the fuming `H₂SO₄` is labeled as 112%, we need to find out how much actual `H₂SO₄` is present in the 12.5 grams of the solution.

\[ \text{Mass of } H₂SO₄ = \left(\frac{112}{100}\right) \times 12.5 \text{ g} = 14 \text{ g} \]

### Step 2: Calculate the number of moles of `H₂SO₄`
Next, we need to calculate the number of moles of `H₂SO₄` using its molar mass. The molar mass of `H₂SO₄` is 98 g/mol.

\[ \text{Moles of } H₂SO₄ = \frac{\text{Mass of } H₂SO₄}{\text{Molar mass of } H₂SO₄} = \frac{14 \text{ g}}{98 \text{ g/mol}} = \frac{14}{98} \text{ moles} \]

### Step 3: Determine the number of moles of `H⁺` produced
Since `H₂SO₄` dissociates completely in solution, it produces 2 moles of `H⁺` ions for every mole of `H₂SO₄`. Therefore, we can calculate the moles of `H⁺` produced.

\[ \text{Moles of } H⁺ = 2 \times \text{Moles of } H₂SO₄ = 2 \times \frac{14}{98} = \frac{28}{98} \text{ moles} \]

### Step 4: Calculate the molar concentration of `H⁺`
The molar concentration (C) is calculated using the formula:

\[ C = \frac{\text{Number of moles}}{\text{Volume of solution in liters}} \]

Given that the volume of water is 100 liters, we can now find the concentration of `H⁺`.

\[ C_{H⁺} = \frac{\frac{28}{98}}{100} = \frac{28}{9800} = \frac{2}{700} \text{ mol/L} \]

### Final Answer
The molar concentration of `H⁺` in the resultant solution is:

\[ C_{H⁺} = \frac{2}{700} \text{ mol/L} \]